Finding $\lim\limits_{n \to \infty}(\sin(1)\cdot \sin(2) \cdot \ldots \cdot \sin(n))$ Problem: Find $\lim_{n \to \infty}(\sin(1)\cdot \sin(2) \cdot \ldots \cdot \sin(n))$.
My idea: I suppose it to be $0$, but how might I go about proving this?
 A: We have $|\sin(n)| \le 1$ for each integer $n \ge 1$. 
Consider two consecutive terms $\sin(n),\sin(n+1)$. Their arguments are "$n$ radians" and "$n+1$ radians". When reduced modulo $2\pi$, for at least one of those two arguments, its distances in the circle from $\pi/2$ and $3\pi/2$ both exceed $\frac{1}{2}$ radian. It follows that one of $|\sin(n)|$ or $|\sin(n+1)|$ is less than 
$$\sin(\frac{\pi-1}{2}) = .877… < .9
$$
Therefore, 
$$|\sin(n) \sin(n+1)| < .9
$$
and so 
$$|\sin(1) \cdot \sin(2) \cdot … \cdot \sin(2n+(\text{0 or 1}))| < (.9)^n
$$
A: Since the integers are equidistributed modulo $2\pi$, there are infinitely many $n$ such that $-1/2<\sin(n)<1/2$. Combining this with the fact that $\sin(n)$ is always at most $1$ in absolute value, we find that the limit is $0$.
Equidistribution in fact tells us the exact convergence rate of $|\sin(1)\sin(2)\cdots\sin(n)|$. Taking logarithms, we have that each of the $n$ integers are equally likely to appear from anywhere in the circle, and so $$\sum_{k=1}^{n}\log(|\sin(k)|)=(1+o(1))\frac{n}{2\pi}\int_{0}^{2\pi}\log(|\sin(x)|)dx$$ 
$$=-n\log2+o(n),$$ and so $$\biggr|\prod_{k=1}^n \sin(k)\biggr|=\left(\frac{1}{2}\right)^{n(1+o(1))}.$$  
A: Hint. Let $n_k = \lfloor \pi k \rfloor$ so that $(n_k)$ is strictly increasing and $|n_k - \pi k| \leq 1$. Can you check that
$$ |\sin n_k| \leq \sin 1 < 1 ?$$
How this leads to the conclusion?
A: $$\sin(k)\sin(k+1)=\frac{1}{2} [\cos(1)-\cos(2k+1)]$$
Now use the fact that $.5 < \cos(1) < .6$ to conclude that 
$$\left| \sin(k) \sin(k+1)\right| < 0.8$$
and 
$$\left| \sin(1)\cdot \sin(2) \cdot \ldots \cdot \sin(n) \right| \leq 0.8^{\frac{n}{2}}$$
A: Consider that in every interval of the form $[\pi n-\frac{1}2,\pi n+\frac{1}2]$ there is at least one integer. Call the set of integers within such an interval $S$ and note that, as there are infinitely many such (disjoint) intervals, $S$ is infinite. However, in such intervals $|\sin(x)|<\sin(\frac{1}2)<1$. If we let $r=\sin(\frac{1}2)$, this, together with the fact that $|\sin(x)|\leq 1$, it follows that a product including the sines of $k$ members of $S$ has absolute value at most $r^k$. Since $k$ goes to infinity, the products can be constrained to any neighborhood of $0$, hence the product converges to $0$.
One can easily make this argument as rigorous as they desire.
A: First off observe that $|\sin n| \leq 1$ for all $n$. Then can you show that there are infinitely many $n$ which are a bounded distance from 1 (i.e. at most equal to $1 - \epsilon$ for some $\epsilon > 0$)?
Then it follows that $$\prod_{n \in \mathbb N} |\sin n \,| \leq \prod_{n \in \mathbb{N}} (1 - \epsilon) = \lim_{n \rightarrow \infty} (1-\epsilon)^n$$
